A) \[p-\sqrt{{{p}^{2}}+{{q}^{2}}}:p+\sqrt{{{p}^{2}}+{{q}^{2}}}\]
B) \[p+\sqrt{{{p}^{2}}-{{q}^{2}}}:p-\sqrt{{{p}^{2}}-{{q}^{2}}}\]
C) \[p:q\]
D) \[p+\sqrt{{{p}^{2}}+{{q}^{2}}}:p-\sqrt{{{p}^{2}}+{{q}^{2}}}\]
E) \[q+\sqrt{{{p}^{2}}-{{q}^{2}}}:q-\sqrt{{{p}^{2}}-{{q}^{2}}}\]
Correct Answer: B
Solution :
According to question, \[\frac{x+y}{2\sqrt{xy}}=\frac{p}{q}\] \[\Rightarrow \] \[\frac{{{(x+y)}^{2}}}{4xy}=\frac{{{p}^{2}}}{{{q}^{2}}}\] ?..(i) On subtracting both sides by 1, we get \[\Rightarrow \] \[\frac{{{(x-y)}^{2}}}{4xy}=\frac{{{p}^{2}}-{{q}^{2}}}{{{q}^{2}}}\] ?. (ii) From Eqs. (i) and (ii), we get \[{{\left( \frac{x+y}{x-y} \right)}^{2}}=\frac{{{p}^{2}}}{{{p}^{2}}-{{q}^{2}}}\] \[\Rightarrow \] \[\frac{x+y}{x-y}=\frac{p}{\sqrt{{{p}^{2}}-{{q}^{2}}}}\] \[\Rightarrow \] \[\frac{2x}{2y}=\frac{p+\sqrt{{{p}^{2}}-{{q}^{2}}}}{p-\sqrt{{{p}^{2}}-{{q}^{2}}}}\] (by componendo-dividendo rule) \[\therefore \]\[x:y=(p+\sqrt{{{p}^{2}}-{{q}^{2}}}):(p-\sqrt{{{p}^{2}}-{{q}^{2}}})\]
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